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### Section 2.8 : Limits at Infinity, Part II

6. For $$\displaystyle f\left( x \right) = \frac{{{{\bf{e}}^{ - 7x}} - 2{{\bf{e}}^{3x}} - {{\bf{e}}^x}}}{{{{\bf{e}}^{ - x}} + 16{{\bf{e}}^{10x}} + 2{{\bf{e}}^{ - 4x}}}}$$ evaluate each of the following limits.

1. $$\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)$$
2. $$\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)$$
Hint : Remember that you’ll need to factor the term in the denominator that is causing the denominator to go to infinity fastest from both the numerator and denominator in order to evaluate this limit.

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a $$\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)$$ Show Solution

The exponentials with the negative exponents are the only terms in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the most negative exponent in the denominator (because it will be going to infinity fastest) from both the numerator and denominator to evaluate this limit.

\begin{align*}\mathop {\lim }\limits_{x \to \, - \infty } \frac{{{{\bf{e}}^{ - 7x}} - 2{{\bf{e}}^{3x}} - {{\bf{e}}^x}}}{{{{\bf{e}}^{ - x}} + 16{{\bf{e}}^{10x}} + 2{{\bf{e}}^{ - 4x}}}} & = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{{\bf{e}}^{ - 4x}}\left( {{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{7x}} - {{\bf{e}}^{5x}}} \right)}}{{{{\bf{e}}^{ - 4x}}\left( {{{\bf{e}}^{3x}} + 16{{\bf{e}}^{14x}} + 2} \right)}}\\ & = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{7x}} - {{\bf{e}}^{5x}}}}{{{{\bf{e}}^{3x}} + 16{{\bf{e}}^{14x}} + 2}} = \frac{{\infty - 0 - 0}}{{0 + 0 + 2}} = \require{bbox} \bbox[2pt,border:1px solid black]{\infty }\end{align*}

b $$\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)$$ Show Solution

The exponentials with the positive exponents are the only terms in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the most positive exponent in the denominator (because it will be going to infinity fastest) from both the numerator and denominator to evaluate this limit.

\begin{align*}\mathop {\lim }\limits_{x \to \, \infty } \frac{{{{\bf{e}}^{ - 7x}} - 2{{\bf{e}}^{3x}} - {{\bf{e}}^x}}}{{{{\bf{e}}^{ - x}} + 16{{\bf{e}}^{10x}} + 2{{\bf{e}}^{ - 4x}}}} & = \mathop {\lim }\limits_{x \to \, \infty } \frac{{{{\bf{e}}^{10x}}\left( {{{\bf{e}}^{ - 17x}} - 2{{\bf{e}}^{ - 7x}} - {{\bf{e}}^{ - 9x}}} \right)}}{{{{\bf{e}}^{10x}}\left( {{{\bf{e}}^{ - 11x}} + 16 + 2{{\bf{e}}^{ - 14x}}} \right)}}\\ & = \mathop {\lim }\limits_{x \to \, \infty } \frac{{{{\bf{e}}^{ - 17x}} - 2{{\bf{e}}^{ - 7x}} - {{\bf{e}}^{ - 9x}}}}{{{{\bf{e}}^{ - 11x}} + 16 + 2{{\bf{e}}^{ - 14x}}}} = \frac{{0 - 0 - 0}}{{0 + 16 + 0}} = \require{bbox} \bbox[2pt,border:1px solid black]{0}\end{align*}